Problem: Rewrite the equation by completing the square. $x^{2}+2x-48 = 0$ $(x + $
Solution: Begin by moving the constant term to the right side of the equation. $x^2 + 2x = 48$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $2$, half of it would be $1$, and squaring it gives us ${1}$. $x^2 + 2x { + 1} = 48 { + 1}$ We can now rewrite the left side of the equation as a squared term. $( x + 1 )^2 = 49$